(10) Suppose you prepared a solution in a 500mL volumetric flask which contains
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(10) Suppose you prepared a solution in a 500mL volumetric flask which contains 5.7080g of Ca(OH)2 and water as the solvent Express the concentration of the calcium hydroxide solution in terms of molarity, ppm and wino. Molar mass for calcium hydroxide 74.093g/mol. (11 a If you now transfer 10ml of the solution from question #10 intoa 100-mL volumetric flask using a volumetric pipet and dilute it to the mark with distilled water, calculate the final concentration of the diluted solution in molarity. Calculate the % inherent error for this diluted solution. Refer to the absolute errors listed on the next page when solving for %inherent error. You will assume there is no error in the molar mass. Refer to the last page of the study guide for the appropriate equation. Please keep in mind that on the actual exam, you will be expected to know the equations and how to use them on your own. (11b) If you now transfer 15ml of the solution from question #10 into a 100-mL volumetric flask usi a 10-mL volumetric pipet and dilute it to the mark with distilled water, calculate the final c inherent error for this diluted solution. Refer to the absolute errors listed on the next ng a 5-mL and oncentration and the % page when solving for ll assume there is no error in the molar mass. Refer to the last page of the study guide for %inherent error. You wi the appropr iate equation. Please keep in mind that on the actual exam, you will be expected to know the equations and how to use them on your own. (11c) What is 1 ppm of a kilometer?Explanation / Answer
10) Molar mass of Ca(OH)2 = 74.093 g/mol.
Mass of Ca(OH)2 taken = 5.7080 g.
Mole(s) of Ca(OH)2 taken = (mass of Ca(OH)2 taken)/(molar mass of Ca(OH)2 taken) = (5.7080 g)/(74.093 g/mol) = 0.0770 mole.
Volume of the solution = 500 mL = (500 mL)*(1 L/1000 mL) = 0.5 L.
Molar concentration of Ca(OH)2 = (moles of Ca(OH)2/volume of solution in L) = (0.0770 mole)/(0.5 L) = 0.154 mol/L = 0.154 M (ans).
We know that 1 ppm solution = 1 mg/L solution; therefore, the concentration of Ca(OH)2 solution in mg/L = (5.7080 g)*(1000 mg/1 g)/(0.5 L) = 11416 mg/L = 11416 ppm (ans).
We have 5.7080 g Ca(OH)2 in 500 mL solution; hence, %w/v of the solution = (5.7080 g)/(500 mL)*100 = 1.1416% (ans).
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